Brachistochrone with velocity - still a cycloid

In summary: ENIn summary, the Brachistochrone problem states that if a particle has an initial velocity v_0 \neq 0, the path of minimum time is still a cycloid. This is proved using the conservation of energy equation and solving the resulting differential equation, which leads to the conclusion that the solution is a cycloid.
  • #1
76
0

Homework Statement


Brachistochrone problem: if the particle is given an initial velocity [tex]v_0 \neq 0[\tex] , show that the path of minimum time is still a cycloid.

Homework Equations


Conservation of energy:
[tex]\frac{1}{2}mv^2-mgy=\frac{1}{2}mv_0^2[\tex]

The Attempt at a Solution


I know how to start the problem, but in the end have to solve the differential equation:
[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-v_0^2-2gy}{v_0^2+2gy}}[\tex]
which I can't solve. Any ideas and hints would be greatly appreciated!
 
Physics news on Phys.org
  • #2
Anyone with a decent solution?
 
  • #4
Grand said:

Homework Statement


Brachistochrone problem: if the particle is given an initial velocity [tex]v_0 \neq 0[/tex] , show that the path of minimum time is still a cycloid.

Homework Equations


Conservation of energy:
[tex]\frac{1}{2}mv^2-mgy=\frac{1}{2}mv_0^2[/tex]

The Attempt at a Solution


I know how to start the problem, but in the end have to solve the differential equation:
[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-v_0^2-2gy}{v_0^2+2gy}}[/tex]
which I can't solve. Any ideas and hints would be greatly appreciated!
You have to put at forward slash to close a latex instruction ie: /tex

Think of the initial velocity as the result of a vertical change in position of a height y0 along this path from a starting point of 0. Then you just have:

[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-2gy0-2gy}{2gy0+2gy}}[/tex]

which reduces to:

[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-2g(y0+y)}{2g(y0+y)}}[/tex]

[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-2gy'}{2gy'}}[/tex]

the solution to which is a cycloid.

AM
 
  • #5




The Brachistochrone problem is a classic example of the application of the principle of least action in physics. The principle states that the path taken by a particle between two points in space is the one that minimizes the action, which is the integral of the Lagrangian over time. In this case, the Lagrangian is the difference between the kinetic and potential energies of the particle.

Using the conservation of energy equation provided in the homework, we can rewrite the differential equation as follows:

[tex]\frac{dy}{dx}=\sqrt{\frac{2g(y_0-y)-v_0^2}{v_0^2+2gy}}[\tex]

Where [tex]y_0[\tex] is the initial height of the particle and [tex]y[\tex] is the height at a given point on the path. This is a separable differential equation, and can be solved using standard techniques.

However, it is important to note that even though the initial velocity [tex]v_0[\tex] is non-zero, it does not affect the shape of the path. This is because the principle of least action still applies, and the path that minimizes the action is still the cycloid. The initial velocity only affects the time taken to travel along the path, but not the shape of the path itself.

To demonstrate this, we can take the derivative of the differential equation with respect to [tex]v_0[\tex]:

[tex]\frac{\partial}{\partial v_0}\left(\frac{dy}{dx}\right)=\frac{1}{2}\frac{1}{\sqrt{\frac{2g(y_0-y)-v_0^2}{v_0^2+2gy}}}\left(\frac{-4gy}{(v_0^2+2gy)^2}-\frac{v_0}{\sqrt{2g(y_0-y)-v_0^2}}\right)[\tex]

Setting this equal to zero and solving for [tex]v_0[\tex], we get:

[tex]v_0=\sqrt{2g(y_0-y)}[\tex]

This means that for any given point on the path, the initial velocity [tex]v_0[\tex] is determined solely by the difference in height between the starting point [tex]y_0[\tex] and the current point
 

1. What is a Brachistochrone curve?

A Brachistochrone curve is a mathematical curve that represents the path of fastest descent between two points in a gravitational field. It was first studied by mathematician Johann Bernoulli in 1696.

2. What is the significance of adding velocity to a Brachistochrone curve?

Adding velocity to a Brachistochrone curve changes the shape of the curve from a cycloid to a trochoid, which allows for a faster descent. This is because the velocity helps to counteract the effects of gravity, allowing the object to reach the bottom of the curve more quickly.

3. How is a Brachistochrone curve with velocity calculated?

A Brachistochrone curve with velocity is calculated using the principles of calculus, specifically the Brachistochrone problem which involves finding the shortest time between two points in a gravitational field. The resulting curve is a cycloid with a modified radius and frequency to account for the added velocity.

4. What are the real-world applications of a Brachistochrone curve with velocity?

A Brachistochrone curve with velocity has many practical applications, such as designing roller coasters, optimizing flight paths for airplanes, and creating efficient water slides. It can also be used in physics and engineering to study the motion of objects in a gravitational field.

5. Are Brachistochrone curves with velocity always the fastest path between two points?

No, Brachistochrone curves with velocity are not always the fastest path between two points. They are only the fastest path if the starting and ending points are at the same height and there are no other external forces acting on the object. In other scenarios, a different curve or path may be faster.

Suggested for: Brachistochrone with velocity - still a cycloid

Back
Top